Entanglement never at first order

When two initially uncorrelated quantum systems interact through a weak coupling, no entanglement is generated at first order in the coupling constant. This is a useful and very easy to prove fact that I haven’t seen pointed out anywhere, although I assume someone has. I’d love a citation reference if you have one.

Suppose two systems \mathcal{A} and \mathcal{B} evolve under U = \exp(- i H t) where the Hamiltonian coupling them is of the form

(1)   \begin{align*} H=H_A + H_B + \epsilon H_I, \end{align*}

with H_A = H_A \otimes I_B and H_B = I_A \otimes H_B as usual. We’ll show that when the systems start out uncorrelated, \vert \psi^0 \rangle = \vert \psi_A^0 \rangle \otimes \vert \psi_B^0 \rangle, they remain unentangled (and therefore, since the global state is pure, uncorrelated) to first order in \epsilon. First, note that local unitaries cannot change the entanglement, so without loss of generality we can consider the modified unitary

(2)   \begin{align*} U' = e^{+i H_A t} e^{+i H_B t} e^{-i H t} \end{align*}

which peels off the unimportant local evolution of \mathcal{A} and \mathcal{B}. Then the Baker–Campbell–Hausdorff formula gives

(3)   \begin{align*} U' = e^{+i H_A t} e^{+i H_B t} e^{-i (H_A + H_B) t} e^{-i \epsilon H_I t}  e^{Z_2} e^{Z_3} \cdots \end{align*}

where the first few Z‘s are given by

(4)   \begin{align*} Z_2 &= \frac{(-i t)^2}{2} [H_A+H_B,\epsilon H_I] \\ Z_3 &= \frac{(-i t)^3}{12} \Big( [H_A+H_B,[H_A+H_B,\epsilon H_I]]-  [\epsilon H_I,[H_A+H_B,\epsilon H_I]] \Big) \\ Z_4 &= \cdots. \end{align*}

The key feature here is that every commutators in each of the Z‘s contains at least one copy of \epsilon H_I, i.e. all the Z‘s are at least first order in \epsilon. That allows us to write

(5)   \begin{align*} U' = e^{-i \epsilon H'_I t} \big(1 + O(\epsilon^2) \big) \end{align*}

for some new H'_I that is independent of \epsilon. Then we note just that a general Hamiltonian cannot produce entanglement to first order:

(6)   \begin{align*} \rho_A &= \mathrm{Tr}_B \left[ U' \vert \psi^0 \rangle \langle \psi^0 \vert {U'}^\dagger \right] \\ &=  \vert \psi'_A \rangle \langle \psi'_A \vert + O(\epsilon^2) \end{align*}


(7)   \begin{align*} \vert \psi'_A \rangle &= \left( I - i \epsilon t \langle \psi^0_B  \vert H_I' \vert  \psi^0_B \rangle \right) \vert \psi^0_A \rangle . \end{align*}

This is potentially a very important (negative) result when considering decoherence detection of very weakly coupled particles. If the coupling is so small that terms beyond first order are negligible (e.g. relic neutrinos), then there is no hope of being sensitive to any decoherence.

Of course, non-entangling (unitary) effect may be important. Another way to say this result is: Two weakly coupled systems act only unitarily on each other to first order in the coupling constant.

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